(Analysis by Freddie Tang and Nick Wu)

It can be proven that the optimal tuition Bessie should charge will always be one of the $c_i$ values (for if not, you could always increase tuition slightly without changing the set of cows who pay). So all we have to do is iterate over these values and see how much money each one makes. To do this, we must first sort these $c_i$ values to determine how many cows will be willing to pay, and then iterate from the lowest to the highest, keeping a count of the number of cows willing to pay, (after a $c_i$ value is visited, the cow with this value will no longer be able to pay the tuition so the count will be decremented). Remember to use $\texttt{long long}$'s because the answer could be up to $10^5\cdot 10^6=10^{11}$.

Freddie Tang's C++ code:

#include <bits/stdc++.h>

using namespace std;

int main() {
int n; cin >> n;
long long maxTuition[n];
for (int i = 0; i < n; i++) {
cin >> maxTuition[i];
}

sort(maxTuition, maxTuition + n);

long long maxMoney = 0, bestTuition = 0;
int numberOfCowsWillingToAttend = n;
for (int i = 0; i < n; i++) {
long long setTuitionToIthCow = maxTuition[i]*numberOfCowsWillingToAttend;
if (setTuitionToIthCow > maxMoney) {
maxMoney = setTuitionToIthCow;
bestTuition = maxTuition[i];
}

numberOfCowsWillingToAttend--; // The ith cow isn't willing to pay any more tuition
}

cout << maxMoney << " " << bestTuition << endl;

return 0;
}


Spencer Compton's Java code:

import java.util.Arrays;
import java.util.Scanner;

public class cowschool {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] a = new int[n];
for(int i = 0; i<n; i++) {
a[i] = in.nextInt();
}
long best = -1L;
int tuit = 0;
Arrays.sort(a);
for(int i = n-1; i>=0; i--) {
long cur = (long)(n-i)*(long)a[i];
if (cur >= best) {
best = cur;
tuit = a[i];
}
}
System.out.println(best+" "+tuit);
}
}


There is an alternate solution that does not involve sorting the cows. For each $p_i$, we can maintain the number of cows that are willing to pay up to $p_i$. If we loop over tuition values going from $10^6$ down to $1$, we can maintain a running total of how many cows are willing to pay that much tuition and track the maximum amount of money that Bessie can make.

Nick Wu's Python code:

MAX_TUITION = 1000000
n = int(input())
tuitionToCow =  * (MAX_TUITION + 1)
for x in input().split():
tuitionToCow[int(x)] += 1
maxMoney = 0
bestTuition = 0
currentCows = 0
for tuition in range(MAX_TUITION, 0, -1):
currentCows += tuitionToCow[tuition]
if tuition * currentCows >= maxMoney:
maxMoney = tuition * currentCows
bestTuition = tuition
print(maxMoney, bestTuition)